Since 7 is not a perfect square, $\displaystyle\sqrt{{7}}\usepackage{cmbright}\Huge\$ is irrational. However, all integers are rational, so $\displaystyle\sqrt{{7}}\usepackage{cmbright}\Huge\$ cannot be an integer. The correct answer is $\displaystyle\sqrt{{{7}}}\usepackage{cmbright}\Huge\$ belongs to the set of irrational and real numbers.